Increasing Xor

The core move is to stop thinking about the operations procedurally and ask what each position can become.

For a query $[l,r]$, suppose the final sequence is

$b_l,b_{l+1},...,b_r$.

Position $i$ can only be affected by operations using source indices $j<=i$. Also, values at those source indices themselves only come from even earlier original values. So no value from the right can ever leak left. Therefore

$b_i$ must belong to the XOR-span of $a_l,a_{l+1},...,a_i$.

This is necessary. It is also sufficient, which is the part that makes the problem click.

Build the final values from right to left. When setting position $i$, positions $l,...,i-1$ are still untouched originals, because we have not processed them yet. If the desired linear combination uses $a_i$, leave it; otherwise zero position $i$ with the operation $i=i$. Then XOR in exactly the earlier original values needed for the combination. Later changes to those earlier positions do not affect position $i$ anymore.

So the query is equivalent to this cleaner problem:

Can we choose a strictly increasing sequence $b_l<b_{l+1}<...<b_r$ such that $b_i$ belongs to $S_i$, where $S_i=span(a_l,...,a_i)$ over GF(2)?

Greedy

At position $i$, say the previous chosen value is $last$. The best choice is the smallest value in $S_i$ that is strictly greater than $last$.

If some valid solution chooses a larger value, replacing it with the smaller valid one cannot hurt any future position. It only lowers the bar for the rest of the sequence. So the greedy choice is forced, not just a cute guess.

Now we need fast operations on a XOR-span:

• count how many span values are $<=x$;
• find the $k$-th smallest value in the span.

With these, if the current span is $S$ and current value is $last$, then the value after taking $k$ consecutive greedy choices from the same fixed span is

$kth(countLeq(last)+k)$.

If that index is outside the span size, the span ran out of usable values.

XOR basis order statistics

Store a standard XOR basis, one vector per highest set bit. Walk bits from high to low.

For $kth(k)$, every pivot bit splits the remaining span values into two equal halves: values with this bit 0 and values with this bit 1. To produce sorted order, decide which half contains the $k$-th value, and XOR the basis vector if needed to force the current bit.

For $countLess(x)$, again walk high to low. Whenever $x$ has bit 1 at a controllable pivot bit, all values with a smaller prefix at that bit contribute half of the remaining combinations. If a bit is not controllable, then the current forced prefix either already became smaller/larger than $x$, so the answer is decided immediately.

Then $countLeq(x)=countLess(x+1)$.

Only change points matter

Fix a left endpoint $l$. The spans

$span(a_l), span(a_l,a_{l+1}), ..., span(a_l,...,a_n)$

only change when the next array value is linearly independent of the current span. Since numbers are below $2^{20}$, the rank is at most 20, so there are at most 20 change points.

Between two change points, the allowed span is constant. Suppose at a change point index $p$ the greedy value becomes $v$, and the current rank is $d$. The span has exactly $2^d$ values. The number of additional positions we can fill before this same span runs out is

$2^d-countLeq(v)$.

So for each $l$, we simulate only those few change points and compute $maxR[l]$, the farthest right endpoint that still works.

At the next change point $idx$:

• if $idx-1$ is beyond the current reachable extension, we fail before the new basis vector arrives;
• otherwise jump through the unchanged-span gap;
• add $a_{idx}$ to the basis;
• choose the next greedy value from the enlarged span;
• update how far this enlarged span can carry us.

Maintaining change points while moving left

Process $l$ from right to left. Keep the change points for the current starting index.

When adding $a_l$:

• If it is independent of the full current basis, it simply becomes the new first change point; all old change points stay valid.
• If it is dependent, some old change points may disappear. Rebuild the change-point list from only the old change points plus $l$, in increasing order. This is enough because any old non-change point was already dependent on earlier change points, and adding another earlier vector cannot make it independent. There are at most 21 candidates, so this rebuild is cheap.

After all $maxR[l]$ are computed, each query is just:

$[l,r]$ is YES iff $r<=maxR[l]$.

That is the whole trick: prefix-span reachability, greedy increasing choices, and only 20 meaningful span changes per left endpoint. The problem looks cursed until those three pieces line up; then it is just a sharp XOR-basis implementation.